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lara31 [8.8K]
3 years ago
11

A ball is dropped from rest out of a high window in a tall building for 5 seconds. Assuming we ignore air resistance and assume

upwards to be positive. A) What will be the final velocity of the ball B) What is the height of the building if it hits the ground after those 5 seconds.
Physics
2 answers:
EleoNora [17]3 years ago
6 0

Answer:

final velocity=-49m/s and height=122.5

Explanation:

u=0,t=5s,v=?

v=u+gt

v=0+(-9.8)5

v=-49m/s

h=ut+1/2gt*2

h=0(5)±1/2(-9.8)5*2

h=-122.5 or 122.5

but height=122.5 because height cannot be negative

lara [203]3 years ago
5 0

Answer:

what is the upwards force?

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a boy throws a ball straight up into the air. it reaches its highest point after 4 seconds.how fast was the ball going when it l
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Explanation:

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3 0
3 years ago
A 57 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. The acceleration of
kari74 [83]

Answer:

Her altitude as she crosses the bar, h₂ is approximately 6.1 m

Explanation:

The given parameters of the motion of the pole vaulter are;

The mass of the pole vaulter, m = 57 kg

The speed with which the pole vaulter is running, u = 11 m/s

The speed of the pole vaulter when she crosses the bar, v = 1.1 m/s

The acceleration due to gravity, g = 9.8 m/s²

From the total mechanical energy, M.E. equation, we have;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy of the motion = m·g·h

K.E. = The kinetic energy of the motion = 1/2·m·v²

By the principle of conservation of energy, we have;

The change (loss) in kinetic energy, ΔK.E. = The change (gain) in potential energy, ΔP.E.

ΔK.E. = 1/2·m·(v² - u²)

ΔP.E. = m·g·(h₂ - h₁)

Where;

h₁ = The ground level = 0 m

h₂ = The altitude with which she crosses the bar

∴ 1/2·m·(v² - u²) = m·g·(h₂ - h₁)

(h₂ - h₁) = (v² - u²)/(2·g) = (11² - 1.1²)/(2·9.8) = 6.11173469388

h₂ = 6.11173469388 + h₁ = 6.11173469388 + 0 = 6.11173469388

h₂ = 6.11173469388

Her altitude as she crosses over the bar, h₂ ≈ 6.1 m.

3 0
3 years ago
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